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First order DE with exponential

  1. Sep 22, 2009 #1
    I want to solve a first order differential equation on the form

    [tex] y'+y=e^{x}[/tex]

    I want to separate the variables, but not sure how....Thanks!
     
  2. jcsd
  3. Sep 22, 2009 #2

    Mute

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    You can't separate the variables directly. The way to solve it is to multiply the whole equation by [itex]e^x[/itex]. You'll then notice that since [itex]d (e^x)/dx = e^x[/itex], the right hand side is just the product-rule expansion of [itex]d(e^xy)/dx[/itex]. So, you have

    [tex]\frac{d}{dx}(e^x y) = e^{2x}.[/tex]

    Now you can separated the variables. You may wonder why I multiplied by e^x. In general you would multiply by some unknown function [itex]\mu(x)[/itex], which you then chose by demanding that you can write the left hand side as [itex]d(\mu y)/dx[/itex]. This requires setting the coefficient of y (after multiplication by the [itex]\mu[/itex]) equal to [itex]d\mu/dx[/itex]. In this case it gives [itex]d\mu/dx = \mu[/itex], which gives the exponential.

    In general, the factor that you multiply by will be

    [tex]\mu(x) = \exp\left[\int dx~(\mbox{coefficient of}~y)\right].[/tex]

    Notes: this trick only works for first order linear equations, you may not get a [itex]\mu(x)[/itex] which you can express in terms of elementary functions (but it still gives you a solution), and the arbitrary constant of integration in the above expression doesn't matter (it will cancel out in the end).
     
  4. Sep 22, 2009 #3
    Thank you very much!
     
  5. Sep 22, 2009 #4

    LCKurtz

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    You have already received one reply using linear methods. An even easier method for this is to observe that it is a constant coefficient equation with characteristic equation r+1=0. So the complementary solution is yc = Ce(-x) and you can find a particular solution yp by undetermined coefficients. Then the general solution is:

    y = yc + yp
     
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